Check sibling questions


Transcript

Question 8 Find the middle terms in the expansions of (๐‘ฅ/3+9๐‘ฆ)^10 Number of terms n = 10 Since n is even. There will be one middle term Middle term = ๐‘›/2 + 1 = 10/2 + 1 = 5 + 1 = 6th term. Hence we need to find 6th term. i.e. T6 We know that general term of (a + b)n is Tr + 1 = nCr (a)n-r . (b)r Finding 6th term i.e. T6 = T5 + 1 , Hence r = 5 Putting n = 5 , a = ๐‘ฅ/3 & b = 9y and r = 1 in (1) T5 + 1 = 10C5 (๐‘ฅ/3)^(10 โˆ’ 5) . (9y)5 T6 = 10C5 (๐‘ฅ/3)^5 . (9)5 . y5 = 10!/5!(10 โˆ’ 5)! . (๐‘ฅ/3)^5. (32) 5 . y5 = 10!/5!5! . ๐‘ฅ5/35 . 310 . y5 = (10 ร— 9 ร— 8 ร— 7 ร— 6 ร— 5!)/(5 ร— 4 ร— 3 ร— 2 ร— 1 ร—5!) . x5 .310/35 . y5 = (10 ร— 9 ร— 8 ร— 7 ร— 6)/(5 ร— 4 ร— 3 ร— 2 ร— 1 ) . x5 . 35 . y5 = (10 ร— 9 ร— 8 ร— 7 ร— 6 ร—35)/(5 ร— 4 ร— 3 ร— 2 ร— 1 ) . x5 . y5 = 61236 x5 y5 Hence, middle term is 61236 x5 y5

  1. Chapter 7 Class 11 Binomial Theorem
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo